\(\int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx\) [655]

Optimal result

Integrand size = 27, antiderivative size = 129 \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) \sqrt {\cos ^2(c+d x)}}+\frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) \sqrt {\cos ^2(c+d x)}} \]

[Out]

a*cos(d*x+c)*hypergeom([-5/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/(1+n)/(cos(d*x+c)^2)^(1/
2)+a*cos(d*x+c)*hypergeom([-5/2, 1+1/2*n],[1/2*n+2],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(2+n)/(cos(d*x+c)^2)^(1/2
)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2917, 2657} \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {a \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}} \]

[In]

Int[Cos[c + d*x]^6*Sin[c + d*x]^n*(a + a*Sin[c + d*x]),x]

[Out]

(a*Cos[c + d*x]*Hypergeometric2F1[-5/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)
*Sqrt[Cos[c + d*x]^2]) + (a*Cos[c + d*x]*Hypergeometric2F1[-5/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c +
d*x]^(2 + n))/(d*(2 + n)*Sqrt[Cos[c + d*x]^2])

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \cos ^6(c+d x) \sin ^n(c+d x) \, dx+a \int \cos ^6(c+d x) \sin ^{1+n}(c+d x) \, dx \\ & = \frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) \sqrt {\cos ^2(c+d x)}}+\frac {a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09 \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) (1+\sin (c+d x)) \left ((2+n) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+(1+n) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )}{d (1+n) (2+n) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2} \]

[In]

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^n*(a + a*Sin[c + d*x]),x]

[Out]

(a*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*(1 + Sin[c + d*x])*((2 + n)*Hypergeometric2F1[-5/2,
(1 + n)/2, (3 + n)/2, Sin[c + d*x]^2] + (1 + n)*Hypergeometric2F1[-5/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*
Sin[c + d*x]))/(d*(1 + n)*(2 + n)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)

Maple [F]

\[\int \left (\cos ^{6}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )d x\]

[In]

int(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c)),x)

[Out]

int(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c)),x)

Fricas [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

integral((a*cos(d*x + c)^6*sin(d*x + c) + a*cos(d*x + c)^6)*sin(d*x + c)^n, x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)**n*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)*sin(d*x + c)^n*cos(d*x + c)^6, x)

Giac [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)*sin(d*x + c)^n*cos(d*x + c)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^6\,{\sin \left (c+d\,x\right )}^n\,\left (a+a\,\sin \left (c+d\,x\right )\right ) \,d x \]

[In]

int(cos(c + d*x)^6*sin(c + d*x)^n*(a + a*sin(c + d*x)),x)

[Out]

int(cos(c + d*x)^6*sin(c + d*x)^n*(a + a*sin(c + d*x)), x)